Question

The distance of the plane 2x - 3y + 4z = 6 from the origin is equal to

• A. $\frac{2}{\sqrt{29}}$
• B. $\frac{3}{\sqrt{29}}$
• C. $\frac{5}{\sqrt{29}}$
• D. $\frac{6}{\sqrt{29}}$

Answer 1

The correct option is D $\frac{6}{\sqrt{29}}$

If the normal form of the plane is $lx+my+nz=d$ where l,m and n are the direction cosines of the planes normal, then d is the distance of the plane from the origin.
Here 2,-3,4 are the direction ratios. Hence, the required distance is
$d=\frac{6}{\sqrt{2^2+(-3{)}^2+4^2}}=\frac{6}{\sqrt{29}}$